9

Well, this is the most common thing for which mathematicians make fun of physicists. Because we don't bother to cancel out derivatives, and we "NEVER" check if we can imply some rule on our equations. The thing is, that almost all functions, which can appear in nature or real life systems are, in most times, continuous and differentiable. There ...


6

It is true, that in nature there is only one true independent variable, time. All others are "pseudo-independent". They are variables humans bless as independent in order to answer what-if scenarios and to establish mathematical models of systems byways of separation of variables. The common term for these "pseudo-independent" quantities ...


5

You are correct that you cannot (globally) write velocity as a function of distance. For example, as one commenter has already mentioned, throw a ball directly up in the air and wait for it to come down. When the ball is at height $h$ on the way up, it has a positive (upward directed) velocity. When it is at the same height $h$ on the way down, it has a ...


4

Yes. The key aspect in your scenario is: they have the same initial velocity. In your addendum „if yes then how“ you seem to overlook that it takes much more force to accelerate the 100kg stone to same velocity. But when they have the same velocity after you let loose they follow the same trajectory, meaning they reach the same height. There is, however, one ...


4

The difference and the derivative is the same thing, but in the limit that the difference becomes "very small". So here is what I like you to do: Let's assume that the position changes with time according to $s = s(t) = t^3$. Draw this function. [Sidemark: Let's assume that we always use SI units. Including the units in the defining equations ...


4

No! Uniform motion is motion in which velocity is constant and acceleration is 0. Uniform acceleration : Motion in which acceleration is constant , therefore velocity is increasing or decreasing or even just the direction is changing for example a uniform circular motion Note: Every uniform motion is having uniform acceleration but vice-versa does not always ...


4

If you write a position vector using polar coordinates like $$\vec{u}(t) = r(t) \left( \hat{x}\cos\phi(t) + \hat{y} \sin\phi(t) \right)$$ then the magnitude of the squared velocity is $$ \begin{eqnarray} ||\dot{\vec{u}}||^2 & = & ||\dot{r} \left( \hat{x}\cos\phi + \hat{y} \sin\phi \right) + r \dot{\phi}\left( -\hat{x}\sin\phi + \hat{y} \cos\phi \...


4

$\frac{d\vec v}{ds}$ isn't as well-behaved (and so not as useful) as $\frac{d\vec v}{dt}$. This can be seen as follows:$$\frac{d\vec v}{ds}=\frac{d\vec v}{dt}\frac{dt}{ds}.$$ So, $$\frac{d\vec v}{ds}=\frac1v\frac{d\vec v}{dt}.$$ in which $v$ is the body's speed. We see that $\frac{d\vec v}{ds}$ is undefined when the body is at rest and not accelerating (in ...


3

Change in velocity with respect to time has the useful property that its value stays the same when you switch to a coordinate system that is in constant motion, relative to the original one. Change in velocity with respect to distance does not. It's just acceleration (as we know it) divided by the magnitude of the velocity in the local coordinate system.


3

It's a consequence of expressing the speed, $v$, in polar coordinates: $$v = \dot{r} \hat{r} + r \dot{\phi} \hat{\phi} \implies v^2 = \dot{r}^2 + r^2 \dot{\phi}^2.$$ If you want to make intuitive sense of what either term means, think of how position is described in polar coordinates: the radial vector denotes how far away from the origin an object is, ...


2

A simple intuitive explanation of sensitivity is that it is a measure of how 'volatile' the function is to increments in it's inputs. For example, consider the square function $$ f(x) = x^2$$ Suppose, I nudge the input by some quantity $'h'$ $$ f(x+h) = (x+h)^2 = x^2 +2xh + h^2$$ The nudge in the out put due the corresponding nudge in input is exactly given ...


2

$|\vec{v}|$ is the norm of the vector $\vec{v}$, and is a scalar value. If $\vec{v}$ is velocity, $|\vec{v}|$ is speed. $\frac{\mathrm{d}}{\mathrm{d}t}\vec{v} = \vec{a}$, the acceleration vector. When we say "tangential acceleration", the direction is "the tangential direction" $\frac{\mathrm{d}}{\mathrm{d}t} |\vec{v}_t| = |\vec{a}_t|$ is ...


2

You appear to be missing the concept of static friction and thinking only of kinetic friction. Kinetic friction is the friction between two objects in relative motion, also called slipping. This friction force has a magnitude $F_{kinetic}=\mu_k N$ where $N$ is the magnitude of the normal force and $\mu_k$ is the coefficient of kinetic friction. The direction ...


2

Your statements do seem to apply to a rolling object. If there is acceleration, the friction provides the torque required for the corresponding angular acceleration.


2

Let me be specific here. Speed is not the magnitude of velocity but instantaneous speed is the magnitude of instantaneous velocity. As pointed out by Brain Stroke Patient, this happens in the limiting case as $ lim \Delta t \rightarrow 0$. In that case, $|\Delta\vec{x}|\rightarrow D$. I think your confusion is that distance travelled and displacement aren't ...


2

The sketches below should help make this less mysterious.


2

imagine you and 100 other people each have a 1kg stone. (assume you can all throw your stone at the same velocity). You throw your stone up and it reaches a certain height. Then the other 100 people throw their stone up at the same time. What happens ? 100kg worth of stone reaches the same height as your 1kg worth of stone. Just because that separate 100kg ...


2

The answer is yes, they reach the same maximum height, provided we include only the gravitational force (no air effects). The reason can be seen in different ways. Equations of motion. Acceleration in a constant gravitational field is independent on the mass ($m a = F$, but $F$ is proportional to $m$). And only acceleration rules the maximum height. ...


2

Forces acting on each point of the sphere (or any object) due to gravity are the same everywhere when the object is in free fall. Therefore, the net external torque is $0$


1

We know (since Galileo's experiments around 1600) that the velocity of a free falling body constantly increases by $9.8$ m/s every second. Therefore it makes sense to define acceleration as the change of velocity per time. Then the movement of all bodies falling free near the earth can be fully described by just one and the same constant number. Of course ...


1

The careful mathematics goes like this: The rate of change of the speed of the particle is given by $$ \frac{dv}{dt} = \frac{d}{dt}\sqrt{\vec{v}\cdot\vec{v}}. $$ Using the chain and product rules of differentiation, we get $$ \frac{dv}{dt} = \frac{1}{2\sqrt{\vec{v}\cdot\vec{v}}}\frac{d}{dt}\vec{v}\cdot\vec{v} = \frac{1}{2v}\left(\frac{d\vec{v}}{dt}\cdot\vec{...


1

$a_t=d|v_{t}|/dt$ This only gives the magnitude of the tangential acceleration, overall tangential acceleration is a vector quantity.


1

$\vec{v} = \Delta \vec{x}/\Delta t$ for infinitesimally small displacement and time. In that limit D is equal to $\Delta x$. If the displacement isn't infinitesimal then you have the average velocity and that need not be equal to, in magnitude, to the average speed.


1

The short answer is that the flow coming into a stagnation point gets pushed to the side, rather than piling up against the fluid ahead of it. The simplest way to see this mathematically is via the Cauchy-Riemann equations,which were derived from complex-analysis and apply for 2-D functions satisfying Laplace's equation (including inviscid flows in fluid ...


1

The velocity change is calculated as the area under the graph. Here it is simply a triangle, so the particle has an increase in velocity by $20\frac{m}{s}$ in the first 20 seconds. That means, it will only be stationary, if it had a velocity of $-20\frac{m}{s}$ beforehand. However, in the setup it is given that the particle had a positive velocity. Thus, the ...


1

It is important to keep in the back of your mind that $\theta$ is really a function of time $\theta(t)$ and that using this function, one can find the velocity (not speed) of particle $A$ at time $t$: $v(\cos \theta(t), \sin \theta(t))$. Ultimately, we can plug and chug with this velocity vector and get a differential equation of $\theta$; however, this ...


1

No, uniform motion implies constant speed, whereas uniform acceleration means constant acceleration.


1

Derivative has a formal and exact meaning and relies on continuity of your position with respect to time for start. This is why it is exact. If you look close enough the deltas in locations are straight lines if you have continuity.


1

let $\dot{x}=u$ and 'a' is spatial acceleration. $U= \gamma(1,u)$ Acceleration, $A = dU/d\tau = \gamma dU/dt = \gamma(\dot{\gamma},\dot{\gamma}u+a\gamma)$ --(1) as $dt/d\tau = \gamma$ $\dot{\gamma} = \sqrt{1-u^2}$ $\implies$ $\dot{\gamma} = \gamma^3ua$ Plug this in (1) and you will get 4-acceleration. If you say that for constant 4-acceleration, each ...


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